Answer:
Once we got
[tex]x=-1[/tex]
[tex]y=3[/tex]
[tex]z=2[/tex]
[tex]\boxed{\text{The sum is 4}}[/tex]
Step-by-step explanation:
Given the linear system:
[tex]\begin{cases} 2x + 3y-z = 5 \\ x- 3y + 2z = -6 \\ 3x + y - 4z = -8 \end{cases}[/tex]
Let's solve it using matrices. I will use Cramer's rule
[tex]M=\left[\begin{array}{ccc}2&3&-1\\1&-3&2\\3&1&-4\end{array}\right][/tex]
Considering determinant as D.
[tex]D=\begin{vmatrix}2&3&-1\\1&-3&2\\3&1&-4\\\end{vmatrix}=40[/tex]
[tex]M_x = \left[\begin{array}{ccc}5&3&-1\\-6&-3&2\\-8&1&-4\end{array}\right] \implies D_x = \begin{vmatrix}5&3&-1\\-6&-3&2\\-8&1&-4\\\end{vmatrix}=-40[/tex]
[tex]M_y = \left[\begin{array}{ccc}2&5&-1\\1&-6&2\\3&-8&-4\end{array}\right] \implies D_y = \begin{vmatrix}2&5&-1\\1&-6&2\\3&-8&-4\\\end{vmatrix}=120[/tex]
[tex]M_z = \left[\begin{array}{ccc}2&3&5\\1&-3&-6\\3&1&-8\end{array}\right] \implies D_z= \begin{vmatrix}2&3&5\\1&-3&-6\\3&1&-8\\\end{vmatrix}=80[/tex]
So, we have
[tex]$x=\frac{D_x}{D} =\frac{-40}{40}=-1 $[/tex]
[tex]$y=\frac{D_y}{D} =\frac{120}{40}=3$[/tex]
[tex]$z=\frac{D_z}{D} =\frac{80}{40}=2 $[/tex]
